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            <td
 style="text-align: center; vertical-align: top; color: rgb(0, 0, 102);"><big><span
 style="font-weight: bold;">About HAP: Cohomology With Twisted
Coefficients<br>
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            <td style="text-align: right; vertical-align: top;"><a
 href="aboutGraphsOfGroups.html"><small style="color: rgb(0, 0, 102);">next</small></a><br>
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      <td
 style="vertical-align: top; background-color: rgb(255, 255, 255); text-align: left;">The
cohomology of a group G with coefficients in a ZG-module A is defined
as:<br>
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 style="background-color: rgb(204, 255, 255); text-align: center; vertical-align: top;"><br>
            <br>
            <span style="color: rgb(0, 0, 102);">H</span><sup
 style="color: rgb(0, 0, 102);">n</sup><span
 style="color: rgb(0, 0, 102);">(G,A) =</span><br>
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 style="text-align: center; vertical-align: top; background-color: rgb(204, 255, 255);">
            <div style="text-align: center;"><span
 style="color: rgb(0, 0, 102);"></span><span
 style="color: rgb(0, 0, 102);"><br>
Ker( Hom<sub>ZG</sub>(R<sub>n</sub>,A) → </span><span
 style="color: rgb(0, 0, 102);">Hom<sub>ZG</sub>(R<sub>n+1</sub>,A) )</span><span
 style="color: rgb(0, 0, 102);"></span><br>
            <hr style="width: 100%; height: 2px;"> <span
 style="color: rgb(0, 0, 102);"></span></div>
            <div style="text-align: center;"><span
 style="color: rgb(0, 0, 102);">Image( </span><span
 style="color: rgb(0, 0, 102);">Hom<sub>ZG</sub>(R<sub>n-1</sub>,A) → </span><span
 style="color: rgb(0, 0, 102);">Hom<sub>ZG</sub>(R<sub>n</sub>,A)<br>
            <br>
            </span></div>
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      <br>
When the abelian group underlying A is free of rank n we can encode A
as a group homomorphism A:G → GL<sub>n</sub>(Z).<br>
      <br>
When G is a permutation group of degree n the free abelian group Z<sup>n</sup>
admits a canonical G-action defined by <br>
      <br>
      <div style="text-align: center;">g·(x<sub>1</sub>, x<sub>2</sub>,
... , x<sub>n</sub>) = (x<sub>g'(1) , xg'(2)</sub> , ... , x<sub>g'(n))
      <br>
      </div>
      <br>
where g'=g-1, for g in G and xi in Z. This
 style="font-style: italic;">canonical permutation module</span> A can
be constructed for any permutation group G using the HAP command <span
 style="font-family: helvetica,arial,sans-serif;">PermToMatrixGroup()</span>.
For example:<br>
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 style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
G:=AlternatingGroup(5);;<br>
      <br>
gap> A:=PermToMatrixGroup(G,5);<br>
[ (1,2,3,4,5), (3,4,5) ] -><br>
[ [ [ 0, 1, 0, 0, 0 ], [ 0, 0, 1, 0, 0 ], [ 0, 0, 0, 1, 0 ],<br>
      [ 0, 0, 0, 0, 1 ], [ 1, 0, 0, 0, 0 ] ],<br>
  [ [ 1, 0, 0, 0, 0 ], [ 0, 1, 0, 0, 0 ], [ 0, 0, 0, 1, 0 ],<br>
      [ 0, 0, 0, 0, 1 ], [ 0, 0, 1, 0, 0 ] ] ]<br>
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      <td
 style="background-color: rgb(255, 255, 255); vertical-align: top;">The
following commands show that:<br>
      <ul>
        <li>the 6th cohomology of the
alternating group G=A<sub>5</sub> with coefficients in its
5-dimensional canonical permutation module A is  H<sup>6</sup>(G,A)
= Z<sub>2</sub>+Z<sub>6</sub>.</li>
        <li>The 3rd cohomology of the even subgroup B<sup>+ </sup>of
the 5-string Braid group, again with coefficients in the permutation
module A (considered as a B<sup>+</sup>-module via the quotient
homomorphism B<sup>+</sup> → A<sub>5</sub>) is H<sup>3</sup>(B<sup>+</sup>,A)
= Z<sub>2</sub>+Z<sub>6</sub>+Z<sup>3</sup>.</li>
      </ul>
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 style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
Alt5:=AlternatingGroup(5);;<br>
gap> A:=PermToMatrixGroup(SymmetricGroup(5),5);;<br>
gap> R:=ResolutionFiniteGroup(Alt5,7);;<br>
gap> TR:=HomToIntegralModule(R,A);;<br>
gap> Cohomology(TR,6);<br>
[ 2, 6 ]<br>
      <br>
      <br>
gap> D:=[[1,[2,3]],[2,[3,3]],[3,[4,3]]];;<br>
gap> R:=ResolutionArtinGroup(D,10);;<br>
gap> Brd5:=R!.group;; Brd5Gens:=GeneratorsOfGroup(Brd5);;<br>
gap>
ImGens:=[Image(A,(1,2)),Image(A,(2,3)),Image(A,(3,4)),Image(A,(4,5))];;<br>
gap> B:=GroupHomomorphismByImages(Brd5,Image(A),Brd5Gens,ImGens);;<br>
gap> EvBrd5:=EvenSubgroup(Brd5);;<br>
gap> S:=ResolutionSubgroup(R,EvBrd5);;<br>
gap> TS:=HomToIntegralModule(S,B);;<br>
gap> Cohomology(TS,3);<br>
[ 2, 6, 0, 0, 0 ]<br>
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      <td
 style="text-align: left; background-color: rgb(255, 255, 255); vertical-align: top;">A
group G can act non-trivially on the integers Z. For example, a
permutation group G can act on Z according to the formula <br>
      <div style="text-align: center;">g.n= -n    if g
is an odd permutation,<br>
      </div>
      <div style="text-align: center;"> g.n=  n   
if g is an even permutation.<br>
      <div style="text-align: left;"><br>
The following commands show that, with this twisted action of S<sub>6</sub>
on Z, we have third twisted integral homology  H<sub>3</sub>(S<sub>6</sub>,Z)=Z<sub>2</sub>+Z<sub>10</sub>
.<br>
      </div>
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      <td
 style="vertical-align: top; background-color: rgb(255, 255, 204);">gap> 
G:=SymmetricGroup(6);;<br>
gap>  R:=ResolutionFiniteGroup(G,4);;<br>
gap>  C:=TensorWithTwistedIntegers(R,SignPerm);;<br>
gap>  Homology(C,3);<br>
[ 2, 10 ]<br>
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      <td
 style="vertical-align: top; background-color: rgb(255, 255, 255);">With
the analogous twisted action of S<sub>6</sub> on Z<sub>5</sub>, the
following commands show that the twelvth homology  is H<sub>12</sub>(S<sub>6</sub>,Z<sub>5</sub>)=Z<sub>5</sub>
. (The calculation relies on the fact that H<sub>n</sub>(G,Z<sub>p</sub>) 
is equal to its p-primary part H<sub>n</sub>(G,Z<sub>p</sub>)<sub>(p)</sub
.) <br>
      </td>
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    <tr>
      <td
 style="vertical-align: top; background-color: rgb(255, 255, 204);">gap> 
G:=SymmetricGroup(6);;<br>
gap> 
P:=SylowSubgroup(G,5);;<br>
gap>  R:=ResolutionFiniteGroup(P,15);;<br>
gap>  F:=function(R);return
TensorWithTwistedIntegersModP(R,5,SignPerm);end;;<br>
gap>  PrimePartDerivedFunctor(G,R,F,12);<br>
[ 5 ]<br>
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Messung V0.5
C=92 H=98 G=94

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